Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $x = \dfrac{4y + 8}{y^2 + 4y + 4} \times \dfrac{-3y^2 + 39y - 90}{y - 10} $
First factor out any common factors. $x = \dfrac{4(y + 2)}{y^2 + 4y + 4} \times \dfrac{-3(y^2 - 13y + 30)}{y - 10} $ Then factor the quadratic expressions. $x = \dfrac {4(y + 2)} {(y + 2)(y + 2)} \times \dfrac {-3(y - 10)(y - 3)} {y - 10} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac {4(y + 2) \times -3(y - 10)(y - 3) } { (y + 2)(y + 2) \times (y - 10)} $ $x = \dfrac {-12(y - 10)(y - 3)(y + 2)} {(y + 2)(y + 2)(y - 10)} $ Notice that $(y + 2)$ and $(y - 10)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac {-12(y - 10)(y - 3)\cancel{(y + 2)}} {\cancel{(y + 2)}(y + 2)(y - 10)} $ We are dividing by $y + 2$ , so $y + 2 \neq 0$ Therefore, $y \neq -2$ $x = \dfrac {-12\cancel{(y - 10)}(y - 3)\cancel{(y + 2)}} {\cancel{(y + 2)}(y + 2)\cancel{(y - 10)}} $ We are dividing by $y - 10$ , so $y - 10 \neq 0$ Therefore, $y \neq 10$ $x = \dfrac {-12(y - 3)} {y + 2} $ $ x = \dfrac{-12(y - 3)}{y + 2}; y \neq -2; y \neq 10 $